动态规划--代码随想录

判断子序列

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        dp = [[0] * (len(t)+1) for _ in range(len(s)+1)]
        for i in range(1, len(s)+1):
            for j in range(1, len(t)+1):
                if s[i-1] == t[j-1]:
                    dp[i][j] = dp[i-1][j-1] + 1
                else:
                    dp[i][j] = dp[i][j-1]
        if dp[-1][-1] == len(s):
            return True
        return False

不同的子序列

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        dp = [[0] * (len(t)+1) for _ in range(len(s)+1)]
        for i in range(len(s)):
            dp[i][0] = 1
        for j in range(1, len(t)):
            dp[0][j] = 0
        for i in range(1, len(s)+1):
            for j in range(1, len(t)+1):
                if s[i-1] == t[j-1]:
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
                else:
                    dp[i][j] = dp[i-1][j]
        return dp[-1][-1]

两个字符串的删除操作

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        dp = [[0] * (len(word2)+1) for _ in range(len(word1)+1)]
        for i in range(len(word1)+1):
            dp[i][0] = i
        for j in range(len(word2)+1):
            dp[0][j] = j
        for i in range(1, len(word1)+1):
            for j in range(1, len(word2)+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j-1] + 2, dp[i-1][j] + 1, dp[i][j-1] + 1)
        return dp[-1][-1]

编辑距离

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        dp = [[0] * (len(word2)+1) for _ in range(len(word1)+1)]
        for i in range(len(word1)+1):
            dp[i][0] = i
        for j in range(len(word2)+1):
            dp[0][j] = j
        for i in range(1, len(word1)+1):
            for j in range(1, len(word2)+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1
        return dp[-1][-1]

回文子串

class Solution:
    def countSubstrings(self, s: str) -> int:
        dp = [[False] * len(s) for _ in range(len(s))]
        result = 0
        for i in range(len(s)-1, -1, -1): #注意遍历顺序
            for j in range(i, len(s)):
                if s[i] == s[j]:
                    if j - i <= 1: #情况一 和 情况二
                        result += 1
                        dp[i][j] = True
                    elif dp[i+1][j-1]: #情况三
                        result += 1
                        dp[i][j] = True
        return result

最长回文子序列

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        dp = [[0] * len(s) for _ in range(len(s))]
        for i in range(len(s)):
            dp[i][i] = 1
        for i in range(len(s)-1, -1, -1):
            for j in range(i+1, len(s)):
                if s[i] == s[j]:
                    dp[i][j] = dp[i+1][j-1] + 2
                else:
                    dp[i][j] = max(dp[i+1][j], dp[i][j-1])
        return dp[0][-1]